Question

A farmer mixes two brands P and Q of cattle feed. Brand P, costing Rs. 250 per bag, contains 3 units of nutritional element A, 2.5 units of elements B and 2 units of element C. Brand Q costing Rs. 200 per bag contains 1.5 units of nutritional elements A, 11.25 units of element B and 3 units of element C. The minimum requirements of nutrients A, B and C are 18 units, 45 units and 24 units respectively. Determine the number of bags of each brand which should be mixed in order to produce a mixture having a minimum cost per bag? What is the minimum cost of the mixture per bag ?

Answer 1

Let the farmer mixes x bags of brand P and y bags of brand Q. We construct the following table :

$\begin{array}{cccccc}Brands& Numberof& ElementA& ElementB& ElementC& Cost\\ P& x& 3x& 2.5x& 2x& 250x\\ Q& y& 1.5y& 11.25y& 3y& 200y\\ Total& x+y& 3x+1.5y& 2.5x+11.25y& 2x+3y& 250x+200y\\ Minimum& & 18& 45& 24& \\ Requires& & & & & \end{array}$

So, our problem is minimize Z = 250x + 200y ....(i)

Subject to constraints $3x+1.5y\ge 18\iff 2x+y\ge 12$ ......(ii)

$2.5x+11.25y\ge 45\iff 2x+9y\ge 36$ ......(iii)

$2x+3y\ge 24$ ......(iv)

$x\ge 0,y\ge 0$ ......(v)

Firstly, draw the graph of the line 3x + 1.5y = 18

$\begin{array}{ccc}x& 6& 0\\ y& 0& 12\end{array}$

Putting (0, 0) in the inequality 3x + 1.5y $\ge$ 18, we have

$3\times 0+1.5\times 0\ge 18\Rightarrow 0\ge 18$ (which is false)

So, the half plane is away from the origin.

Secondly, draw the graph of the line 2.5x + 11.25y = 45

$\begin{array}{ccc}x& 18& 0\\ y& 0& 4\end{array}$

Putting (0, 0) in the inequality 2.5x + 11.25y $\ge$ 45, we have

$2.5\times 0+11.25\times 0\ge 45\Rightarrow 0\ge 45$ (which is false)

So, the half plane is away from the origin.

Thirdly, draw the graph of the line 2x + 3y = 24

$\begin{array}{ccc}x& 0& 12\\ y& 8& 0\end{array}$

Putting (0, 0) in the inequality 2x + 3y $\ge$ 24, we have

$2\times 0+3\times 0\ge 24\Rightarrow 0\ge 24$ (which is false)

So, the half plane is away from the origin. Since, x, y $\ge$ 0

So, the feasible region lies in the first quadrant.

On solving equations 3x + 1.5y = 18 and 2x + 3y = 24, we get C(3, 6).

Similarly, on solving equations 2.5x + 11.25y = 45 and 2x + 3y = 24, we get B(9, 2).

The corner points of the feasible region are A(18, 0), B(9, 2), C(3, 6) and D(0, 12), The values of Z at these points are as follows:

$\begin{array}{cc}Cornerpoint& Z=250x+200y\\ A(18,0)& 4500\\ B(9,2)& 2650\\ C(3,6)& 1950\to Minimum\\ D(0,12)& 2400\end{array}$

As the feasible region is unbounded, therefore 1950 may or may not be the minimum value of Z.

For this, we draw a graph of the inequality 250x + 200y < 1950 or 5x + 4y < 39 and check, whether the resulting half plane has points in common with the feasible region or not.

It can be seen that the feasible region has no common point with 5x + 4y < 39.

Therefore, the minimum value of Z is 1950 at C(3, 6).

Thus, 3 bags of brand P and 6 bags of brand Q should be used in the mixture to minimize the cost to Rs. 1950.